If $A$ is symmetric, then the matrix exponential $e^{A}$ is positive definite

Question

This is a homework problem, but I feel like I'm struggling with not knowing facts from linear algebra. Apparently this is supposed to be an easy question but I hit a brick wall at the following point. Can anyone give me a tip or tell me if I'm going down the wrong road?

My approach:

Let $A$ be a symmetric $n\times n$ matrix, and then suppose $x$ is a real-valued column vector of dimension $n$ with not all entries equal to $0$. I need to show that $x^Te^{A}x> 0$.

(based on looking up the definition of positive definite on wikipedia)

Then

$\begin{eqnarray*} x^{T}(e^{A})x &=& x^{T}(\sum_{n=0}^{\infty}\frac{A^{n}}{n!})x\\ &=& \sum_{n=0}^{\infty}\frac{x^{T}A^{n}x}{n!}\\ &=& \sum_{n=0}^{\infty}\frac{x^{T}A^{n}x}{n!})\\ \end{eqnarray*}$

As noted below, from here on is incorrect:

$\begin{eqnarray*} &=& \sum_{n=0}^{\infty}\frac{(x^{T}Ax)^{n}}{n!\|x\|^{2(n-1)}})\\ &=& \sum_{n=0}^{\infty}\frac{(x^{T}A^{T}x)^{n}}{n!\|x\|^{2(n-1)}}\\ &=& \sum_{n=0}^{\infty}\frac{((Ax)^{T}x)^{n}}{n!\|x\|^{2(n-1)}}\\ &=& \sum_{n=0}^{\infty}\frac{((Ax)^{T}(x^{T})^{T})^{n}}{n!\|x\|^{2(n-1)}}\\ &=& \sum_{n=0}^{\infty}\frac{((x^{T}Ax)^{T})^{n}}{n!\|x\|^{2(n-1)}}\\ \end{eqnarray*}$

As you can see this brings me no closer to getting $x^Tx$ somewhere, which I may assume is greater than $0$.

So my conclusion is that using only the fact that $A$ is symmetric is not enough. Is there some result about symmetric matrices that I should use?

Answer 1

Since $A$ is symmetric and the operator $\cdot^T\colon A\in \mathcal M_n(\mathbb R)\mapsto A^T\in\mathcal M_n(\mathbb R)$ is continuous (as a linear operator in a finite dimensional vector space), the matrix $e^{A/2}$ is symmetric. Therefore, we have for $x\in\mathbb R^n$: $$x^Te^Ax=x^Te^{A/2}e^{A/2}x=x^T(e^{A/2})^Te^{A/2}x =(e^{A/2}x)^Te^{A/2}x=\lVert e^{A/2}x\rVert^2\geq 0,$$ and since $e^{A/2}$ is invertible, we have the equality if and only if $x=0$, which shows that $e^A$ is positive definite.

Answer 2

I assume that $A$ is a real matrix here. Here is a way that requires some knowledge about linear algebra but no calculation. By spectral theorem $A=U \Lambda U^{-1}$ where $\Lambda$ is the diagonal matrix, the entries of which are the eigenvalues of $A$ (hence non negative), call these $\lambda_i\geq 0$. Take now $e^A$ which is clearly symmetric, its eigenvalues are $e^{\lambda_i} > 0$, and you are done.

Answer 3

If $A$ is symmetric, it is diagonalizable. So write $$A = PDP^{-1}$$ where $D = \text{diag}(\lambda_1, ..., \lambda_n)$ is a diagonal matrix whose entries are the eigenvalues of $A$. Thus $$e^A = Pe^DP^{-1}$$ and it is easy to see that $e^D = \text{diag}(e^{\lambda_1}, ..., e^{\lambda_n})$. But since the exponential is always positive, this means that all the eigenvalues of $e^A$ are positive. Hence $e^A$ is positive definite.